# Calculation of Volumes Using Triple Integrals

The volume of a solid U in Cartesian coordinates xyz is given by

$V = \iiint\limits_U {dxdydz} .$

In cylindrical coordinates, the volume of a solid is defined by the formula

$V = \iiint\limits_U {\rho d\rho d\varphi dz} .$

In spherical coordinates, the volume of a solid is expressed as

$V = \iiint\limits_U {{\rho ^2}\sin \theta d\rho d\varphi d\theta } .$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the volume of a cone of height $$H$$ and base radius $$R$$ (Figure $$1$$).

### Example 2

Find the volume of the ball ${x^2} + {y^2} + {z^2} \le {R^2}.$

### Example 3

Find the volume of the tetrahedron bounded by the planes passing through the points $$A\left( {1,0,0} \right),$$ $$B\left( {0,2,0} \right),$$ $$C\left( {0,0,3} \right)$$ and the coordinate planes $$Oxy,$$ $$Oxz,$$ $$Oyz$$ $$\left({\text{Figure }2}\right).$$

### Example 4

Find the volume of the tetrahedron bounded by the planes

$x + y + z = 5, x = 0, y = 0, z = 0$

(Figure $$4$$).

### Example 1.

Find the volume of a cone of height $$H$$ and base radius $$R$$ (Figure $$1$$).

Solution.

The cone is bounded by the surface $$z = {\frac{H}{R}} \sqrt {{x^2} + {y^2}}$$ and the plane $$z = H$$ (see Figure $$1$$).

Its volume in Cartesian coordinates is expressed by the formula

$V = \iiint\limits_U {dxdydz} = \int\limits_{ - R}^R {dx} \int\limits_{ - \sqrt {{R^2} - {x^2}} }^{\sqrt {{R^2} - {x^2}} } {dy} \int\limits_{\frac{H}{R}\sqrt {{x^2} + {y^2}} }^H {dz} .$

Calculate this integral in cylindrical coordinates that range within the limits:

$0 \le \varphi \le 2\pi ,\;\; 0 \le \rho \le R,\;\; \rho \le z \le H.$

As a result, we obtain (do not forget to include the Jacobian $$\rho$$):

$V = \int\limits_0^R {\rho d\rho } \int\limits_0^{2\pi } {d\varphi } \int\limits_{\frac{H}{R}\rho }^H {dz} .$

Then the volume of the cone is

$V = \int\limits_0^R {\rho d\rho } \int\limits_0^{2\pi } {d\varphi } \int\limits_{\frac{H}{R}\rho }^H {dz} = 2\pi \int\limits_0^R {\rho d\rho } \int\limits_{\frac{H}{R}\rho }^H {dz} = 2\pi \int\limits_0^R {\rho d\rho } \cdot \left[ {\left. z \right|_{z = \frac{H}{R}\rho }^{z = H}} \right] = 2\pi \int\limits_0^R {\rho \left( {H - \frac{H}{R}\rho } \right)d\rho } = 2\pi \int\limits_0^R {\left( {H\rho - \frac{H}{R}{\rho ^2}} \right)d\rho } = 2\pi \left[ {\left. {\left( {\frac{{{\rho ^2}H}}{2} - \frac{{{\rho ^3}H}}{{3R}}} \right)} \right|_{\rho = 0}^{\rho = R}} \right] = 2\pi \left( {\frac{{{R^2}H}}{2} - \frac{{{R^3}H}}{{3R}}} \right) = \frac{{2\pi {R^2}H}}{6} = \frac{{\pi {R^2}H}}{3}.$

### Example 2.

Find the volume of the ball ${x^2} + {y^2} + {z^2} \le {R^2}.$

Solution.

We calculate the volume of the part of the ball lying in the first octant $$\left( {x \ge 0,y \ge 0,z \ge 0} \right),$$ and then multiply the result by $$8.$$ This yields:

$V = 8\iiint\limits_U {dxdydz} = 8\iiint\limits_{U'} {{\rho ^2}\sin \theta d\rho d\varphi d\theta } = 8\int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \int\limits_0^{\frac{\pi }{2}} {\sin \theta d\theta } = 8\int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^{\frac{\pi }{2}}} \right] = 8\int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot \left( { - \cos \frac{\pi }{2} + \cos 0} \right) = 8\int\limits_0^{\frac{\pi }{2}} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot 1 = 8\int\limits_0^{\frac{\pi }{2}} {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^3}}}{3}} \right)} \right|_0^R} \right] = \frac{{8{R^3}}}{3}\int\limits_0^{\frac{\pi }{2}} {d\varphi } = \frac{{8{R^3}}}{3} \cdot \left[ {\left. \varphi \right|_0^{\frac{\pi }{2}}} \right] = \frac{{8{R^3}}}{3} \cdot \frac{\pi }{2} = \frac{{4\pi {R^3}}}{3}.$

As a result, we get the well-known expression for the volume of the ball of radius $$R.$$

### Example 3.

Find the volume of the tetrahedron bounded by the planes passing through the points $$A\left( {1,0,0} \right),$$ $$B\left( {0,2,0} \right),$$ $$C\left( {0,0,3} \right)$$ and the coordinate planes $$Oxy,$$ $$Oxz,$$ $$Oyz$$ $$\left({\text{Figure }2}\right).$$

Solution.

The equation of the straight line $$AB$$ in the $$xy$$-plane (Figure $$3$$) is written as $$y = 2 - 2x.$$ The variable $$x$$ ranges here in the interval $$0 \le x \le 1,$$ and the variable $$y$$ ranges in the interval $$0 \le y \le 2 - 2x.$$

We write the equation of the plane $$ABC$$ in segment form. Since the plane $$ABC$$ cuts the line segments $$1, 2,$$ and $$3,$$ respectively, on the $$x-,$$ $$y-,$$ and $$z-$$axis, then its equation can be written as

$\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1.$

In general case the equation of the plane $$ABC$$ is written as

$6x + 3y + 2z = 6\;\;\text{or}\;\; z = 3 - 3x - \frac{3}{2}y.$

Hence, the limits of integration over the variable $$z$$ range in the interval from $$z = 0$$ to $$z = 3 - 3x - {\frac{3}{2}} y.$$

Now we can calculate the volume of the tetrahedron:

$V = \iiint\limits_U {dxdydz} = \int\limits_0^1 {dx} \int\limits_0^{2 - 2x} {dy} \int\limits_0^{3 - 3x - \frac{3}{2}y} {dz} = \int\limits_0^1 {dx} \int\limits_0^{2 - 2x} {dy} \cdot \left[ {\left. z \right|_0^{3 - 3x - \frac{3}{2}y}} \right] = \int\limits_0^1 {dx} \int\limits_0^{2 - 2x} {\left( {3 - 3x - \frac{3}{2}y} \right)dy} = \int\limits_0^1 {dx} \cdot \Big[ {\left. {\left( {3y - 3xy - \frac{3}{4}{y^2}} \right)} \right|_{y = 0}^{y = 2 - 2x}} \Big] = \int\limits_0^1 {\Big[ {3\left( {2 - 2x} \right) - 3x\left( {2 - 2x} \right) - \frac{3}{4}{{\left( {2 - 2x} \right)}^2}} \Big] dx} = \int\limits_0^1 {\Big[ {6 - 6x - 6x + 6{x^2} - \frac{3}{4}\left( {4 - 8x + 4{x^2}} \right)} \Big] dx} = \int\limits_0^1 {\left( {\color{green}{6} - \color{red}{12x} + \color{blue}{6{x^2}} - \color{green}{3} + \color{red}{6x} - \color{blue}{3{x^2}}} \right)dx} = 3\int\limits_0^1 {\left( {\color{green}{1} - \color{red}{2x} + \color{blue}{x^2}} \right)dx} = 3\left[ {\left. {\left( {x - {x^2} + \frac{{{x^3}}}{3}} \right)} \right|_0^1} \right] = 3 \cdot \left( {\cancel{1} - \cancel{1^2} + \frac{{{1^3}}}{3}} \right) = 1.$

### Example 4.

Find the volume of the tetrahedron bounded by the planes

$x + y + z = 5, x = 0, y = 0, z = 0$

(Figure $$4$$).

Solution.

The equation of the plane $$x + y + z = 5$$ can be rewritten in the form

$z = 5 - x - y.$

By setting $$z = 0,$$ we get

$5 - x - y = 0\;\;\text{or}\;\;y = 5 - x.$

Hence, the region of integration $$D$$ in the $$xy$$-plane is bounded by the straight line $$y = 5 - x$$ as shown in Figure $$5.$$

Representing the triple integral as an iterated integral, we can find the volume of the tetrahedron:

$V = \iiint\limits_U {dxdydz} = \int\limits_0^5 {dx} \int\limits_0^{5 - x} {dy} \int\limits_0^{5 - x - y} {dz} = \int\limits_0^5 {dx} \int\limits_0^{5 - x} {dy} \cdot \left[ {\left. z \right|_0^{5 - x - y}} \right] = \int\limits_0^5 {dx} \int\limits_0^{5 - x} {\left( {5 - x - y} \right)dy} = \int\limits_0^5 {dx} \left[ {\left. {\left( {5y - xy - \frac{{{y^2}}}{2}} \right)} \right|_{y = 0}^{y = 5 - x}} \right] = \int\limits_0^5 {\left[ {5\left( {5 - x} \right) - x\left( {5 - x} \right) - \frac{{{{\left( {5 - x} \right)}^2}}}{2}} \right]dx} = \int\limits_0^5 {\left( {25 - 5x - 5x + {x^2} - \frac{{25 - 10x + {x^2}}}{2}} \right)dx} = \frac{1}{2}\int\limits_0^5 {\left( {25 - 10x + {x^2}} \right)dx} = \frac{1}{2}\left[ {\left. {\left( {25x - \frac{{10{x^2}}}{2} + \frac{{{x^3}}}{3}} \right)} \right|_0^5} \right] = \frac{1}{2}\left( {125 - 5 \cdot 25 + \frac{{125}}{3}} \right) = \frac{{125}}{6}.$

See more problems on Page 2.